Find the matrix that corresponds to reflecting over the vector $\begin{pmatrix} 3 \\ 2 \end{pmatrix}.$
Solution: Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix},$ let $\mathbf{r}$ be the reflection of $\mathbf{v}$ over $\begin{pmatrix} 3 \\ 2 \end{pmatrix},$ and let $\mathbf{p}$ be the projection of $\mathbf{v}$ onto $\begin{pmatrix} 3 \\ 2 \end{pmatrix}.$

Note that $\mathbf{p}$ is the midpoint of $\mathbf{v}$ and $\mathbf{r}.$  Thus, we can use $\mathbf{p}$ to compute the reflection matrix.

[asy]
unitsize(1 cm);

pair D, P, R, V;

D = (3,2);
V = (1.5,2);
R = reflect((0,0),D)*(V);
P = (V + R)/2;

draw((-1,0)--(4,0));
draw((0,-1)--(0,3));
draw((0,0)--D,Arrow(6));
draw((0,0)--V,red,Arrow(6));
draw((0,0)--R,blue,Arrow(6));
draw((0,0)--P,green,Arrow(6));
draw(V--R,dashed);

label("$\mathbf{p}$", P, S);
label("$\mathbf{v}$", V, N);
label("$\mathbf{r}$", R, SE);
[/asy]

From the projection formula,
\begin{align*}
\mathbf{p} &= \operatorname{proj}_{\begin{pmatrix} 3 \\ 2 \end{pmatrix}} \begin{pmatrix} x \\ y \end{pmatrix} \\
&= \frac{\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}} \begin{pmatrix} 3 \\ 2 \end{pmatrix} \\
&= \frac{3x + 2y}{13} \begin{pmatrix} 3 \\ 2 \end{pmatrix} \\
&= \begin{pmatrix} \frac{9x + 6y}{13} \\ \frac{6x + 4y}{13} \end{pmatrix}.
\end{align*}Since $\mathbf{p}$ is the midpoint of $\mathbf{v}$ and $\mathbf{r},$
\[\mathbf{p} = \frac{\mathbf{v} + \mathbf{r}}{2}.\]Then
\begin{align*}
\mathbf{r} &= 2 \mathbf{p} - \mathbf{v} \\
&= 2 \begin{pmatrix} \frac{9x + 6y}{13} \\ \frac{6x + 4y}{13} \end{pmatrix} - \begin{pmatrix} x \\ y \end{pmatrix} \\
&= \begin{pmatrix} \frac{5x + 12y}{13} \\ \frac{12x - 5y}{13} \end{pmatrix} \\
&= \begin{pmatrix} 5/13 & 12/13 \\ 12/13 & -5/13 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.
\end{align*}Thus, the matrix is $\boxed{\begin{pmatrix} 5/13 & 12/13 \\ 12/13 & -5/13 \end{pmatrix}}.$